Tricky Fourier Transform problem for an exponential function

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SUMMARY

The discussion focuses on finding the Fourier transform of the function f(t) = 2πt²e^{-πt²} using the definition \widehat{f}(\nu) = ∫^{∞}_{-∞} f(t)e^{-2πiνt} dt. The correct Fourier transform is identified as (1 - 2πν²)e^{-πν²}. The user initially struggled with the integral, attempting substitutions and integration by parts, but received guidance on correcting their approach by shifting the integral along the complex plane and utilizing integration by parts effectively.

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Homework Statement


find the Fourier transform, using the definition of the Fourier transform [itex]\widehat{f}[/itex]([itex]\nu[/itex])=∫[itex]^{∞}_{-∞}[/itex]f(t)e[itex]^{-2 \pi i \nu t}[/itex]dt, of the function f(t)=2 [itex]\pi[/itex]t[itex]^{2}[/itex]e[itex]^{- \pi t^{2}}[/itex]

Homework Equations



I have the answer:

(1-2[itex]{\pi \nu^{2}}[/itex])e[itex]^{- \pi \nu^{2}}[/itex]

The Attempt at a Solution



After inserting f(t) into the equation for the transform, I added the exponents on the e terms, factored out -[itex]\pi[/itex], and added and subtracted [itex]\nu^{2}[/itex] to get

∫[itex]^{∞}_{-∞}[/itex]2[itex]\pi[/itex]t[itex]^{2}[/itex]e[itex]^{- \pi (t^{2}+2 i \nu t + \nu^{2} - \nu^{2})}[/itex]dt

I then substituted x=t+i[itex]\nu[/itex] to get

∫[itex]^{∞}_{-∞}[/itex]2[itex]\pi[/itex](t+i[itex]\nu[/itex])[itex]^{2}[/itex]e[itex]^{- \pi (x^{2}+ \nu^{2} )}[/itex]dt

at this point I know that I should be able to solve the integral, probably by integrating by parts, but I am really just lost. This seems like such a tricky integral. I thought maybe squaring the equation to get a double integral and then converting to polar coordinates would work but I couldn't get that to work out either. Thanks for the help!
 
Last edited:
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eaglemike said:

Homework Statement


find the Fourier transform, using the definition of the Fourier transform [itex]\widehat{f}[/itex]([itex]\nu[/itex])=∫[itex]^{∞}_{-∞}[/itex]f(t)e[itex]^{-2 \pi i \nu t}[/itex]dt, of the function f(t)=2 [itex]\pi[/itex]t[itex]^{2}[/itex]e[itex]^{- \pi t^{2}}[/itex]

Homework Equations



I have the answer:

(1-2[itex]{\pi \nu^{2}}[/itex])e[itex]^{- \pi \nu^{2}}[/itex]

The Attempt at a Solution



After inserting f(t) into the equation for the transform, I added the exponents on the e terms, factored out -[itex]\pi[/itex], and added and subtracted [itex]\nu^{2}[/itex] to get

∫[itex]^{∞}_{-∞}[/itex]2[itex]\pi[/itex]t[itex]^{2}[/itex]e[itex]^{- \pi (t^{2}+2 i \nu t + \nu^{2} - \nu^{2})}[/itex]dt

I then substituted x=t+i[itex]\nu[/itex] to get

∫[itex]^{∞}_{-∞}[/itex]2[itex]\pi[/itex](t+i[itex]\nu[/itex])[itex]^{2}[/itex]e[itex]^{- \pi (x^{2}+ \nu^{2} )}[/itex]dt

at this point I know that I should be able to solve the integral, probably by integrating by parts, but I am really just lost. This seems like such a tricky integral. I thought maybe squaring the equation to get a double integral and then converting to polar coordinates would work but I couldn't get that to work out either. Thanks for the help!

Your last expression is wrong: it should be [itex]\int_{-\infty + i \nu}^{\infty + i \nu} 2 \pi x^2 e^{-\pi (x^2 + \nu^2)}\, dx ,[/itex], which can be replaced by the same integral from x = -infinity to +infinity along the real x-axis (why)? Now int x^2*exp(-x^2) dx can be attacked using integration by parts.

RGV
 
Last edited:

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